Big Ideas Math: Modeling Real Life, Grade 8
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3. Solving Systems of Linear Equations by Elimination
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Exercise 41 Page 218

Use the information from the exercise to create a system of linear equations. Then, use the Elimination Method to solve it.

90 % gold alloy: 5 grams, 50 % gold alloy: 3 grams

Practice makes perfect
We want to find out how much of each alloy the jeweler should use. Let's create a system of linear equations to help us and then solve it using the Elimination Method. We will use x to represent the mass of the 90 % gold alloy and y to represent the mass of the 50 % gold alloy. Mass of the90 % gold alloy - x Mass of the50 % gold alloy - y From the exercise we know that the jeweler wants to make 8 grams of 18-karat gold. This means that the combined mass of both alloys should be 8. Let's write this fact as an equation. x+ y= 8We also know that 18-karat gold is 75 % gold. To create this alloy, the jeweler will combine two alloys with different content of gold, one with 90 % gold and the other one with 50 % gold. The final product will have the mass of these two alloys combined and its gold content will be 75 %. Remember that we can write percents as decimal numbers. Let's take a look. 0.9* x+ 0.5* y= 0.75( x+ y) Now that we have written two equations, we can combine them into a system of equations. x+y=8 & (I) 0.9x+0.5y=0.75(x+y) & (II) Let's rewrite Equation (II) so that all the variable terms are on one side of the equation. This will make solving the system easier.
0.9x+0.5y=0.75(x+y)
0.9x+0.5y=0.75x+0.75y
0.15x+0.5y=0.75y
0.15x-0.25y=0
Now we are ready to solve the system to find how much of each alloy the jeweler should use. Notice that no pair of like terms has the same or opposite coefficients. Let's multiply Equation (II) by 4 so that the y-terms will have opposite coefficients.
x+y=8 & (I) 0.15x-0.25y=0 & (II)
x+y=8 & (I) 4(0.15x-0.25y)=4(0) & (II)
x+y=8 0.6x-y=0
If we add Equation (II) to Equation (I), we will get an equation in only one variable, x. We also could have multiplied one of the equations by a negative number and then subtracted them — the answer would be the same.
x+y=8 & (I) 0.6x-y=0 & (II)
x+y+( 0.6x-y)=8+( 0) 0.6x-y=0
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Solve for x
x+y+0.6x-y=8+0 0.6x-y=0
1.6x=8 0.6x-y=0
x=5 0.6x-y=0
Let's substitute 5 for x in Equation (II) and solve for y.
x=5 & (I) 0.6x-y=0 & (II)
x=5 0.6( 5)-y=0
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Solve for y
x=5 3-y=0
x=5 -y=-3
x=5 y=3
We found that the jeweler should use 5 grams of the 90 % gold alloy and 3 grams of the 50 % gold alloy.